If $x=\frac{1}{t^2}$ and $y=\frac{1}{t^3}$, then $\frac{d^2y}{dx^2}$ at t = 1 is :
Answer & explanation
Correct answer: option 4
$x=\frac{1}{t^2},y=\frac{1}{t^3}$ [Given]
$\frac{dx}{dt}=-2\frac{1}{t^3}$
$\frac{dy}{dt}=-3\frac{1}{t^4}$
$=\frac{dy}{dx}=\frac{3}{2t}$